Convex Lens Real
Images:
The Geometry

THE SETUP:

In the diagram above, two principle rays are used to locate the
real image that is to the right of the lens and inverted.

TOWARDS AN EQUATION:

In this diagram, several lines are given symbols which are used
to designate their lengths. The chart below defines each.
Typically these would be measured in centimeters or possibly
millimeters. In astronomical systems, meters might be used.

Symbol

Meaning

Ho

Size (height) of object. (o represents object)

Do

Distance from lens to the object

f

focal length (lens to focal point)

Do - f

Distance from first focal point to object
can be negative if object is closer than focal point

Hi

Size (height) of image (i represents image)

Di

Distance from lens to image

Di - f

Distance from second focal point to image
can be negative if object is between focal point and lens

APPROXIMATION:

We consider the thickness of the lens to be small compared to
the distances involved. We think for simplification that the light
rays do all of their bending in the center of the lens rather than
the two sides. This gives us the ability to do some basic geometry
as shown in the diagram which follows.

The lightly shaded triangles on the left form similar triangles
because they share vertical angles. (We assume the object is
perpendicular to the axis as is the lens.) Likewise we get similar
triangles on the right, the more darkly shaded ones.

Because both equations contain the term Ho/Hi, we can
set the remaining terms equal to each other and we get the
equation below.

If we cross-multiply the right hand terms and expand the
binomials, we get the following:

We now subtract f2 from both sides and then
move the terms with f to the left-hand side:

Now factor out the f and move the binomial to the
right-hand side:

Here we invoke a mathematical trick, even though the equation
looks okay (and it is). Take the reciprocal of the entire
equation, and then expand the right-hand side:

There it is! An equation that predicts the image location based
on the object distance and the focal length. Any number of
problems can be derived from this equation, and it will be seen to
be applicable for four different optical devices! (Five if you
stretch things.)

Equation 1

We now go on to find another equation, based on a third light
ray that also works with the final image.

We see that there will be similar triangles if we assume both
the object and image are perpendicular to the axis, and if we
assume the lens is thin and the light doesn't deviate from a
straight line very far. This gives us the following equation:

Equation 2

This equation gives the ratio of image size to object size in
terms of image distance and object distance. This ratio is also
called the magnification, m. It is easy to see that if the
real image is formed further from the lens than the object is
placed, the image will also be larger.

What are some predictions from this
equation?

1. If the object distance is very large, the image distance
will be f, the focal length. ie, the image will be formed one
focal length away from the lens.

Substitute a large value for Do in the first
equation. Make it infinity (mathematicians don't like that
concept, so they often call it GBN, Great Big Number). Now
1/infinity or 1/GBN is zero or almost zero.

2. If the object distance is f, the image distance will be
undefined, or infinite. In practice, the lens will form a set of
parallel light rays that will neither converge or diverge, so no
image is ever formed.

3. The object and image distances will be equal at the point
where they are both 2f.

SIGNS:

For lenses, we develop a sense of signs in our mathematics. It
comes from the direction that light travels through the lens. In
the diagram which follows, note that the positive sense of things
occurs when light starts on one side and converges on the other.